CS609 Assignment No 4 Spring 2012 solution

Assignment                                                                  

Part a)  05 Marks

You have studied in this course that Tracks are the circular division of the disk and the Sectors are the longitudinal division of the disk. A typical track is shown in yellow; a typical sector is shown in blue. A sector contains a fixed number of bytes — for example, 256 or 512.

As per division like above, there is a clear difference in size of sectors on inner and outer tracks which causes under utilization of the larger, outer tracks of the disk. What mechanism is used to improve disk surface area utilization?

Solution:

Many factors affects the HDD storage capacity.
some of these are disks (platters), increase in surface area, density of megnetic media etc etc

Part b) 15 Marks

Suppose, we have a Hard disk having 16 (Read/Write) Heads per Cylinder, and Maximum number of Sectors per Track is 63. It has total 1024 cylinders and Sector size is 512 Bytes. Answer the following questions.

Q#1. Find the LBA address of the CHS address (4, 6, 12)?

Q#2. Find the CHS address of the LBA address 4256?

Q#3. Find the Total Data Storage Capacity of this Hard Disk?

Solution:

LBA Address

LBA address of the CHS address (4, 6, 12)

LBA Address=(c*h’+h)*S’+s-1

c=4, h=6, S=12, H’=16, S’ = 63

=(4*16+6)*63+12-1

=(64+6)63+12-1

=4421

CHS Address

cylender =LBA/(heads per cylender*sectors per track)

LBA=4421, heads per cylender=16, sector per track is 63

So we have

4421/(16*63)

=4.386

Total data storage capacity.

cylinders x heads x sectors x sector size

1024 * 16 * 63 * 512 = 528482304 = 516 MB approx

 

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